Capacitors Connected, Disconnected, and Connected Again

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Homework Statement

Capacitors 1,3, and 3 have capacitances equal to 2.0uF, 4.0 uF, and 6.0uF, respectively. The capacitors are connected in parallel, and the parallel combination is connected across the terminals of a 200-V source. The capacitors are then disconnected from both the voltage source and each other, and are connected to three switches as shown in the image. a) What is the potential difference across each capacitor when switches S1 and S2 are closed but switch S3 remains open? b) After switch S3 is closed, what is the final charge on the lefmost plate of each capacitor? c) Give the final potential difference across each capacitor after switch S3 is closed.

Homework Equations

C = Q/V

Vsys=Cequiv*Qtotal

Series Connection: 1/Cequiv = 1/C1 + 1/C2 + ... + 1/Cn
Parallel Connection: Cequiv = C1 + C2 +...+ Cn
n= number of capacitors under consideration.

The Attempt at a Solution

For a) the potentials are the same, but I think I am not justifying it by the right reasons. My argument is that since all the capacitors were at the same potential when connected in parallel, as they are connected as asked by a there can be no direct change in potential because none of the capacitors are directly attached to one another.
b) I really don't know exactly, and trust me, I'm not trying to slack off but I'm just clueless as to how to approach it. My thoughts is that C1 and C3 are connected in parallel but I don't know how to account for C2 so everything remains unsolved. If they were all in series then they would all end up with the same charge, but that's not the case so I am not sure what to do. Any help is greatly appreciated.

 

[LowlyPion]

Welcome to PF.

When they are in || across the voltage source they are endowed with specific charges from the supply. Q = V*C

What they want you to consider then is what happens when you take these charged capacitors and now place them in a series configuration.

 

[nlightner]

a) When switches S1 and S2 are closed, but switch S3 remains open, the potential difference across each capacitor will be equal to the potential of the source, which is 200V.

b) After switch S3 is closed, the final charge on the leftmost plate of each capacitor will depend on the equivalent capacitance of the circuit. The equivalent capacitance of the circuit can be calculated using the formula for parallel connection: Cequiv = C1 + C2 + C3.

Since C1 and C3 are connected in parallel, their equivalent capacitance will be (2.0uF)*(6.0uF)/(2.0uF+6.0uF) = 1.5uF. This equivalent capacitance will then be connected in series with C2, resulting in a total equivalent capacitance of (1.5uF)*(4.0uF)/(1.5uF+4.0uF) = 1.2uF.

Using the formula Q = CV, we can find the final charge on the leftmost plate of each capacitor: Q = (1.2uF)*(200V) = 240uC.

c) After switch S3 is closed, the final potential difference across each capacitor can be calculated using the formula V = Q/C. The potential difference across C1 will be 240uC/2.0uF = 120V, across C2 will be 240uC/4.0uF = 60V, and across C3 will be 240uC/6.0uF = 40V.